সৃজনশীলঃ ‘10×2−8x+1=0 এবং 2×3−3×2+4x−1=0…’ কুমিল্লা বোর্ড ২০২২ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

এখন. $10x^2-8x+1=0$ ......... (i)

ধরি, (1) নং মূলদ্বয় $\alpha$ এবং $\beta$; যেখানে $\alpha >\beta$

$\therefore \alpha +\beta =-\frac{-8}{10}=\frac{4}{5}$

এবং $\alpha \beta =\frac{1}{10}$

$(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$

$={\left(\frac{4}{5}\right)}^2-4\cdot \frac{1}{10}=\frac{16}{25}-\frac{2}{5}=\frac{16-10}{25}$

$\therefore (\alpha -\beta {)}^2=\frac{6}{25}$

বा, $\alpha -\beta =\sqrt{\frac{6}{25}}=\frac{\sqrt{6}}{5}$

$\therefore (\alpha +\beta )$ এবং $(\alpha -\beta )$ মূলদ্বয়বিশিষ্ট সমীকরণ হবে,

$x^2-\left\{\right(\alpha +\beta )+(\alpha -\beta \left)\right\}x+(\alpha +\beta )(\alpha -\beta )=0$

বা, $X^2-(\frac{4}{5}+\frac{\sqrt{6}}{5})x+\frac{4}{5}\cdot \frac{\sqrt{6}}{5}=0$

বা, $x^2-\frac{4+\sqrt{6}}{5}x+\frac{4\sqrt{6}}{25}=0$

$\therefore 25x^2-5(4+\sqrt{6})x+4\sqrt{6}=0$

নির্ণেয় দ্বিঘাত সমীকরণ $25x^2-5(4+\sqrt{6})x+4\sqrt{6}=0$. 

এখানে, $2x^3-3x^2+4x-1=0$

সমীকরণ (1) এর মূলত্রয় $\alpha ,\beta ,\gamma$

সুতরাং, $\alpha +\beta +\gamma =-\frac{-3}{2}=\frac{3}{2}$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{4}{2}=2$

$\alpha \beta \gamma =-\frac{-1}{2}=\frac{1}{2}$

$\therefore \Sigma {\alpha }^2\beta ={\alpha }^2\beta +{\alpha }^2\gamma +{\beta }^2\alpha +{\beta }^2\gamma +{\gamma }^2\beta +{\gamma }^2\alpha$

$\therefore \Sigma {\alpha }^2\beta ={\alpha }^2\beta +{\alpha }^2\gamma +{\beta }^2\alpha +{\beta }^2\gamma +{\gamma }^2\beta +{\gamma }^2\alpha$

$={\alpha }^2\beta +{\beta }^2\alpha +\alpha \beta \gamma +{\beta }^2\gamma +{\gamma }^2\beta +\alpha \beta \gamma +{\gamma }^2\alpha +{\alpha }^2\gamma +\alpha \beta \gamma -3\alpha \beta \gamma$

বা, $\Sigma {\alpha }^2\beta =\alpha \beta (\alpha +\beta +\gamma )+\beta \gamma (\alpha +\beta +\gamma )+\gamma \alpha (\alpha +\beta +\gamma )-3\alpha \beta \gamma$

বা, $\Sigma {\alpha }^2\beta =(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )-3\alpha \beta \gamma$

বা, $\Sigma {\alpha }^2\beta =\frac{3}{2}\cdot 2-3\cdot \frac{1}{2}=3-\frac{3}{2}=\frac{3}{2}$

$\therefore \Sigma {\alpha }^2\beta =\frac{3}{2}$

নির্ণেয় মান $\frac{3}{2}$.